#2807

Medium

auto_awesomeLinked-list pointer manipulation

LeetCode Problem Workspace

Insert Greatest Common Divisors in Linked List

The problem involves inserting greatest common divisors between adjacent nodes in a linked list.

Problem Statement

You are given the head of a linked list where each node contains an integer. Between every pair of adjacent nodes, insert a new node containing the greatest common divisor (GCD) of the two values.

Return the linked list after performing the insertions. If there is only one node in the list, return the list as is. Ensure to handle lists with varying node counts and values efficiently.

Examples

Example 1

Input: head = [18,6,10,3]

Output: [18,6,6,2,10,1,3]

The 1st diagram denotes the initial linked list and the 2nd diagram denotes the linked list after inserting the new nodes (nodes in blue are the inserted nodes).

We insert the greatest common divisor of 18 and 6 = 6 between the 1st and the 2nd nodes.
We insert the greatest common divisor of 6 and 10 = 2 between the 2nd and the 3rd nodes.
We insert the greatest common divisor of 10 and 3 = 1 between the 3rd and the 4th nodes. There are no more adjacent nodes, so we return the linked list.

Example 2

Input: head = [7]

Output: [7]

The 1st diagram denotes the initial linked list and the 2nd diagram denotes the linked list after inserting the new nodes. There are no pairs of adjacent nodes, so we return the initial linked list.

Constraints

The number of nodes in the list is in the range [1, 5000].
1 <= Node.val <= 1000

Solution Approach

GCD Calculation and Linked List Traversal

To solve this problem, traverse the linked list while calculating the GCD of each adjacent pair of nodes. After calculating the GCD, create a new node with the GCD value and insert it between the current pair of nodes. Move the pointer to the next adjacent pair of nodes and repeat the process.

Efficient Insertion of Nodes

To avoid inefficient insertions, directly manipulate the pointers while traversing. Instead of shifting nodes, just update the 'next' pointers to insert the new nodes with the GCD values. This reduces unnecessary complexity and ensures that the algorithm remains efficient even for larger lists.

Edge Case Handling

Handle edge cases such as when the list has only one node. In this case, no insertions are necessary, so simply return the head. Also, make sure that the GCD calculation handles values at the upper end of the allowed range without causing overflow or performance issues.

Complexity Analysis

Metric	Value
Time	O(n \cdot \log(\min(a,b)))
Space	O(1)

The time complexity of the solution is O(n * log(min(a, b))) due to the GCD calculation for each pair of nodes, where n is the number of nodes in the list and a, b are the node values. The space complexity is O(1) since we are only manipulating the pointers and not using any additional data structures.

What Interviewers Usually Probe

Check if the candidate handles linked list pointer manipulation correctly.
Look for understanding of efficient node insertion without unnecessary shifts.
Test whether the candidate anticipates and handles edge cases like single-node lists.

Common Pitfalls or Variants

Common pitfalls

Failing to efficiently insert nodes, resulting in higher time complexity.
Overlooking edge cases, such as when the list has only one node.
Misunderstanding the GCD calculation, especially with large node values.

Follow-up variants

What if the problem requires you to insert the sum or product of adjacent nodes instead of the GCD?
Modify the problem to handle sorted linked lists or to add other number-theoretic operations.
Adjust the problem to use a doubly linked list instead of a singly linked list.

FAQ

How do I insert nodes with the greatest common divisor in a linked list?

To insert nodes with the greatest common divisor, calculate the GCD for every adjacent pair, create a new node with the GCD value, and insert it between the two nodes.

What is the time complexity of the 'Insert Greatest Common Divisors in Linked List' problem?

The time complexity is O(n * log(min(a, b))) where n is the number of nodes, and a, b are the values of adjacent nodes.

Can the solution handle large linked lists efficiently?

Yes, the solution is designed to handle large lists efficiently by directly manipulating pointers without creating unnecessary copies or shifting nodes.

What is the significance of the greatest common divisor in this problem?

The GCD of adjacent nodes determines the values of the new nodes that must be inserted between each pair, making it an essential step in solving the problem.

How does GhostInterview help with linked list pointer manipulation?

GhostInterview provides step-by-step solutions and highlights key techniques to master linked list pointer manipulation, ensuring you avoid common mistakes and optimize performance.

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Solution

Solution 1: Simulation

We use two pointers $pre$ and $cur$ to point to the current node and the next node respectively. We only need to insert a new node between $pre$ and $cur$. Therefore, each time we calculate the greatest common divisor $x$ of $pre$ and $cur$, we insert a new node with value $x$ between $pre$ and $cur$. Then we update $pre = cur$ and $cur = cur.next$, and continue to traverse the linked list until $cur$ is null.

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def insertGreatestCommonDivisors(
        self, head: Optional[ListNode]
    ) -> Optional[ListNode]:
        pre, cur = head, head.next
        while cur:
            x = gcd(pre.val, cur.val)
            pre.next = ListNode(x, cur)
            pre, cur = cur, cur.next
        return head

Continue Practicing

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